Engineering, 29.10.2019 23:31, myvenus88

The space between the plates of a parallel-plate capacitor is filled with two slabs of linear dielectric material. the first slab has thickness a and the second slab has thickness 2a, so the total distance between the plates is 3a. slab 1 has a dielectric constant 3e0, and slab 2 has a dielectric constant of 2eo0. the free charge on the top plate is σ and the free charge on the bottom plate is -σ. (a) find the electric displacement �++⃗ in each slab. (b) find the electric field �+⃗ in each slab. (c) find the polarization �+⃗ in each slab. (d) find the potential difference between the plates. (e) find the location and amount of all bound charges. (f) now that you know all the charges (free and bound), recalculate the electric field in each slab and confirm your answer to (b).

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The space between the plates of a parallel-plate capacitor is filled with two slabs of linear dielec...